Question

The time spent (in days) waiting for a heart transplant in two states for patients with type A+ blood can be approximated by a normal distribution. Assume u-127, standard deviation=22.6

shortest time spent waiting for a heart is at the top 20% waiting times? Round two decimals as needed

longest time waiting for a heart that would place a patient in the bottom 15% of waiting times? Round two decimals as needed

Answer 1

Part A: calulationg shortest time spent waiting for a heart is at the top 20% waiting times
In terms of probability and decimals top 20% would mean those above 0.8

P(X>t) =0.2 hence, P(X<t) = 0.8
We have to find X, for that we will need Z-score. We can find the z score by looking at standard normal distribution table for the value 0.8 in standard normal distribution table.
The z score comes out to be 0.841

Formula for Z-score
Z = (X-mean) / standard deviation
hence, putting values of mean = 127 and standard deviation = 22.6 in above formula
(X - 127) / 22.6 = 0.841
X = 22.6 * 0.841 + 127
X = 146 (Answer)

Part B. Calculating longest time waiting for a heart that would place a patient in the bottom 15% of waiting times
Using same table Z score will come out to be -1.036
(X - 127) / 22.6 = -1.036
X = 22.6 * (-1.036) + 127
X = 103.58 (Answer)